Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential ( $\mathrm{E}^{\circ}$ ) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $\mathrm{E}^{\circ}$ ( V with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 39-41. $$ \begin{array}{ll} \mathrm{I}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-} & \mathrm{E}^{\circ}=0.54 \\ \mathrm{Cl}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} & \mathrm{E}^{\circ}=1.36 \\ \mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} & \mathrm{E}^{\circ}=1.50 \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} & \mathrm{E}^{\circ}=0.77 \\ \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} & \mathrm{E}^{\circ}=1.23 \end{array} $$
Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and $\mathrm{H}_2 \mathrm{SO}_4$ in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of